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### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Consider the differential equation $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}{\mathbf{=}}{\mathbit{x}}{\mathbf{+}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbf{}}{\mathbit{y}}$⦁ A solution curve passes through the point $\left(1,\frac{\pi }{2}\right)$ . What is its slope at this point?⦁ Argue that every solution curve is increasing for ${\mathbit{x}}{\mathbf{>}}{\mathbf{1}}$ .⦁ Show that the second derivative of every solution satisfies $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{+}}{\mathbf{x}}{}{\mathbf{cos}}{}{\mathbf{y}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{sin}}{}{\mathbf{2}}{\mathbf{y}}{\mathbf{.}}$ ⦁ A solution curve passes through (0,0). Prove that this curve has a relative minimum at (0,0).

⦁ The slope at the point 2.

⦁ Yes, every solution curve is increasing for x > 1.

⦁ The second derivative of every solution satisfies the given equation.

⦁ Yes, the curve has a minimum at (0,0)

See the step by step solution

## 1(a): Find the slope of solution curve at  1,π2

Slope is given by $\frac{\mathbf{dy}}{\mathbf{dx}}$

So, slope of the solution curve at $\left(\mathbf{1}\mathbf{,}\frac{\pi }{\mathbf{2}}\right)$ is

$\frac{\mathbf{dy}}{\mathbf{dx}}\left|\left(\mathbf{1}\mathbf{,}\frac{\pi }{\mathbf{2}}\right)\right\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{sin}\left(\mathbf{1}\mathbf{,}\frac{\pi }{\mathbf{2}}\right)\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{1}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{2}$

Hence, the slope at the point is 2.

## 2(b): Compute  dydx for x > 1

Since, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{y}$ for all $\mathbf{y}\in \mathrm{ℝ}$

Then, $\mathbf{x}\mathbf{+}\mathbf{sin}\mathbf{y}\mathbf{>}\mathbf{1}$ , for all $\mathbf{x}\mathbf{>}\mathbf{1}$.

i.e., $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{>}\mathbf{1}\mathbf{>}\mathbf{0}$ , for all $\mathbf{x}\mathbf{>}\mathbf{1}$, $\mathbf{y}\in \mathrm{ℝ}$.

Hence, from first derivative test, every solution curve is increasing for ${\mathbf{x}}{\mathbf{>}}{\mathbf{1}}$ .

## 3(c): Determine the second derivative.

Here

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{y}$

Differentiate both sides with respect to x

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{y}\left(\frac{\mathbf{dy}}{\mathbf{dx}}\right)\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{+}\mathbf{sin}\mathbf{y}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{cos}\mathbf{y}\mathbf{+}\mathbf{sin}\mathbf{y}\mathbf{cos}\mathbf{y}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{cos}\mathbf{y}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{sin}\mathbf{2}\mathbf{y}$

So, the second derivative of every solution satisfies the given equation.

## 4(d): Find second derivative at (0,0)

Since, $\mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{y}\mathbf{=}\mathbf{0}$ at $\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{\right)}$

we get, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{0}$ at $\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{\right)}$

thus, $\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{\right)}$ is a critical point.

Also, from part (c) , $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{cos}\mathbf{y}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{sin}\mathbf{2}\mathbf{y}$

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{>}\mathbf{0}$ at $\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{\right)}$

From second derivative test, $\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{\right)}$ is a point of relative minimum.

Therefore, the curve has a minimum at $\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{\right)}$